Question

Find the integral of $\frac{x\log x}{{(1+x^2)}^2}$ under the limits $0$ to infinity.

Answer 1

Find the integral of the given function

Given integral:${\int }_0^{\infty }\frac{x\log x}{{(1+x^2)}^2}dx$

Let $I={\int }_0^{\infty }\frac{x\log x}{{(1+x^2)}^2}dx$ and $x=\frac{1}{t}$

So $dx=-\frac{1}{t^2}dt$, for $x=0,t\to \infty$ , for $x\to \infty ,t=0$ and we get

$$\begin{gathered} I={\int }_{\infty }^0\frac{\left(\frac{1}{t}\right)\log \left(\frac{1}{t}\right)\left(-\frac{1}{t^2}\right)}{{\left(1+{\left(\frac{1}{t}\right)}^2\right)}^2}dt \\ \Rightarrow I={\int }_{\infty }^0\frac{\left(\frac{1}{t}\right)\log \left(\frac{1}{t}\right)\left(-\frac{1}{t^2}\right)}{{\left(\frac{t^2+1}{t^4}\right)}^2}dt \\ \Rightarrow I={\int }_{\infty }^0\frac{\left(\frac{1}{t}\right)\left(-\log t\right)\left(-\frac{1}{t^2}\right)}{{\left(\frac{t^2+1}{t^4}\right)}^2}dt\left[\because \log \left(\frac{1}{t}\right)=-\log t\right] \\ \Rightarrow I={\int }_{\infty }^0\frac{t\log t}{{\left(t^2+1\right)}^2}dt \\ \Rightarrow I=-{\int }_0^{\infty }\frac{t\log t}{{\left(1+t^2\right)}^2}dt \\ \Rightarrow I=-I \\ \Rightarrow 2I=0 \\ \Rightarrow I=0 \\ \therefore {\int }_0^{\infty }\frac{x\log x}{{(1+x^2)}^2}dx=0 \end{gathered}$$

Hence, the integral of $\frac{x\log x}{{(1+x^2)}^2}$ under the limits $0$ to infinity is $0$.